(A) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.Initial moles: $N_2O_4 = 0.8 \ mol$,$NO_2 = 0 \ mol$.At equilibrium,let the amount of

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(A) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.Initial moles: $N_2O_4 = 0.8 \ mol$,$NO_2 = 0 \ mol$.At equilibrium,let the amount of $N_2O_4$ reacted be $x \ mol$.Then,$[N_2O_4] = (0.8 - x) \ M$ and $[NO_2] = 2x \ M$ (since volume is $1 \ L$).$K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(2x)^2}{0.8 - x} = 0.00466$.$4x^2 = 0.00466(0.8 - x) \implies 4x^2 + 0.00466x - 0.003728 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x \approx 0.03$.Therefore,$[NO_2] = 2x = 2 \times 0.03 = 0.06 \ M$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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The reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is started by taking $0.8 \ mol$ of $N_2O_4$ in a $1 \ L$ flask. If the equilibrium constant at $298 \ K$ is $0.00466 \ M$,the equilibrium concentration of $NO_2$ will be ........... $M$.

A
$0.06$
B
$0.03$
C
$0.74$
D
$0.36$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$,the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $8.0 \ mol \ L^{-1}$,$3.0 \ mol \ L^{-1}$,and $28.0 \ mol \ L^{-1}$ respectively. The equilibrium constant $(K_c)$ of the reaction is: (in $.66$)

MediumAIPMT 2001

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The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

DifficultAIEEE 2004

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For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ at $25^o C$,the value of $K_c$ is $6.10 \times 10^{-3}$. What is the value of $K_c$ for the equilibrium reaction $NO_{2(g)} \rightleftharpoons 1/2 N_2O_{4(g)}$?

Medium

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For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,at equilibrium,the mixture of steam and hydrogen contains $40\%$ $H_2$ by volume. Find $K_p$ for the reaction.

Medium

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For the reaction $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$, if we write $K_p = K_c(RT)^x$, then $x$ becomes

EasyWBJEE 2009

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The reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ is started by taking $0.8 \ mol$ of $N_2O_4$ in a $1 \ L$ flask. If the equilibrium constant at $298 \ K$ is $0.00466 \ M$,the equilibrium concentration of $NO_2$ will be ........... $M$.

Similar Questions

For the reaction $H_2 + I_2 \rightleftharpoons 2HI$,the equilibrium concentrations of $H_2$,$I_2$,and $HI$ are $8.0 \ mol \ L^{-1}$,$3.0 \ mol \ L^{-1}$,and $28.0 \ mol \ L^{-1}$ respectively. The equilibrium constant $(K_c)$ of the reaction is: (in $.66$)

The equilibrium constant for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{1}{2}O_{2(g)}$ at the same temperature is

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$ at $25^o C$,the value of $K_c$ is $6.10 \times 10^{-3}$. What is the value of $K_c$ for the equilibrium reaction $NO_{2(g)} \rightleftharpoons 1/2 N_2O_{4(g)}$?

For the reaction $SnO_{2(s)} + 2H_{2(g)} \rightleftharpoons 2H_2O_{(g)} + Sn_{(l)}$,at equilibrium,the mixture of steam and hydrogen contains $40\%$ $H_2$ by volume. Find $K_p$ for the reaction.

For the reaction $SO_2 + \frac{1}{2} O_2 \rightleftharpoons SO_3$, if we write $K_p = K_c(RT)^x$, then $x$ becomes

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